|
|
A081808
|
|
Numbers n such that the largest prime power in the factorization of n equals phi(n).
|
|
2
|
|
|
12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
All numbers 3*2^k k>=2 are in the sequence.
Let n=p^k*q where p^k is the largest prime power is the factorization of n and (p,q)=1. If n belongs to the sequence then p^k = phi(n) = (p-1)*p^(k-1)*phi(q), implying that p=2 (since p-1 cannot divide p^k for prime p>2). Then 2 = phi(q), implying that q=3. Therefore the terms are simply the sequence 3*2^n for n=2,3,... - Max Alekseyev, Mar 02 2007
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 3*2^(n+1).
|
|
MATHEMATICA
|
|
|
PROG
|
|
|
CROSSREFS
|
Essentially the same as A007283 = 3*2^n.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|